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is sqrt x an integer?

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is sqrt x an integer?

by heffsz » Thu Dec 11, 2008 7:16 pm
Please help! Thanks.

If x is a positive integer, is sqrt(x) an integer?
(1) sqrt(4x) is an integer
(2) sqrt(3x) is not an integer

[spoiler]OA is A. (1) reasoning is that if sqrt(4x) is an integer than both 4 and x must be the squares of integers. if x is the square of an integer then clearly sqrt(x) is an integer.

what i dont understand in (1) is how we definitely know that both 4 and x must be the squares of integers? 4 we know has 2 as its squared root, but how can we safely say that x is? i feel like im forgetting some rule regarding products and or squares... also if (1) instead read that sqrt(8x) is an integer then following the previously mentioned logic i would reason that both 8 and x must be the squares of integers as well, when we all know 8 is not the square of an integer.

(2) could use some clarification on this as well.[/spoiler]
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by srisl11 » Thu Dec 11, 2008 7:54 pm
We can plug in numbers too for this ...

But I try to avoid plugging in numbers because I fear that I might lose some special cases ...


Given x is an integer
Is sqrt (x) an integer?

st 1 :

sqrt (4x) is an integer

sqrt (4x) = k , Where k is an integer

=> 4x = k^2

x= k^2 / 4
Since x is an integer => k^2 /4 is also an integer
In otherwords k^2 is a mulitple of 4

x= k^2 / 4
taking sqrt on both sides
sqrt (x) = k/2
Since k^2 is a multiple of 4
=> k must be a multiple of 2

=>sqrt(x) = k/2 = integer

st 1 is sufficient

Statement 2

Sqrt (3x) =p (non integer)

sqrt(x) = p/1.73 (may or may not be an integer depending on the value of p)
So Statement 2 is insuff

SO A
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by cramya » Thu Dec 11, 2008 8:19 pm
Pick numbers

Stmt II

sqrt(4x) is an integer

2*sqrt(x) is an integer

x has to be an integer and a perfect square therefore sqrt(x) is an integer

SUFF

Stmt I

sqrt(3x) is not an integer

sqrt(3) * sqrt(x) is not an integer

x =1/2
x=6

Both satisfy not an integer condition when multiplied by sqrt(3)

INSUFF


Choose A)
Last edited by cramya on Thu Dec 11, 2008 9:48 pm, edited 1 time in total.
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by srisl11 » Thu Dec 11, 2008 8:28 pm
cramya wrote:Pick numbers

Stmt II

sqrt(4x) is an integer

2*sqrt(x) is an integer

x has to be an integer and a perfect square

SUFF


Choose A)
There may be people like me who might think that what if sqrt(x) = 3.5
then 2 *3.5 = integer..
But sqrt (x) cannot be any decimal ending with "yyy.5" since x is an integer

Am I right here?
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by srisl11 » Thu Dec 11, 2008 8:30 pm
Thank you Cramya for a simple approach...
:)
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by cramya » Thu Dec 11, 2008 8:32 pm
There may be people like me who might think that what if sqrt(x) = 3.5
then 2 *3.5 = integer..
But sqrt (x) cannot be any decimal ending with "yyy.5" since x is an integer

Am I right here
Sris,
U r right IMO. If x was 1.5 or 3.5 the sqrt(x)^2 i.e. x will not be an integer
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by ronniecoleman » Thu Dec 11, 2008 10:11 pm
IMO A
Admission champion, Hauz khaz
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by heffsz » Fri Dec 12, 2008 7:55 am
thanks for the help everyone!
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by tritrantran » Fri Dec 12, 2008 9:25 am
cramya wrote:Pick numbers

Stmt II

sqrt(4x) is an integer

2*sqrt(x) is an integer

x has to be an integer and a perfect square therefore sqrt(x) is an integer

SUFF

Stmt I

sqrt(3x) is not an integer

sqrt(3) * sqrt(x) is not an integer

x =1/2
x=6

Both satisfy not an integer condition when multiplied by sqrt(3)

INSUFF


Choose A)

What if we pick x = 2 for Statement (1)

sqrt(4x) = int

2*sqrt(x) = int

2*sqrt(2) = int ? False


Then pick x = 1

2*sqrt(1) = int ? True

INSUFFICIENT

Am i missing something?
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by NicoFalcon » Fri Dec 12, 2008 9:32 am
But if you replace x=2 in the original statement:

sqrt4(x) integer, sqrt4(2) is not an integer
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by tritrantran » Fri Dec 12, 2008 12:18 pm
Nevermind, I read the problem wrong. Ok, makes sense now.
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by pbanavara » Fri Dec 12, 2008 11:14 pm
cramya wrote:Pick numbers

Stmt II

sqrt(4x) is an integer

2*sqrt(x) is an integer

x has to be an integer and a perfect square therefore sqrt(x) is an integer

SUFF

Stmt I

sqrt(3x) is not an integer

sqrt(3) * sqrt(x) is not an integer

x =1/2
x=6

Both satisfy not an integer condition when multiplied by sqrt(3)

INSUFF


Choose A)
So the question is "is sqrt(x) an integer" ?? From statement 2 we can say that sqrt(x) is not an integer right ? So shouldn't the answer be D ??? I must be missing something here
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by cramya » Fri Dec 12, 2008 11:28 pm
So the question is "is sqrt(x) an integer" ?? From statement 2 we can say that sqrt(x) is not an integer right ? So shouldn't the answer be D ??? I must be missing something here
U dint miss anything. I should have said x=9 and not put it down as 6

sqrt(3) sqrt(9) is not an intger but sqrt(9) is an integer.

Hence INSUFF.Hope this helps!
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by tritrantran » Sat Dec 13, 2008 8:05 am
cramya wrote:
So the question is "is sqrt(x) an integer" ?? From statement 2 we can say that sqrt(x) is not an integer right ? So shouldn't the answer be D ??? I must be missing something here
U dint miss anything. I should have said x=9 and not put it down as 6

sqrt(3) sqrt(9) is not an intger but sqrt(9) is an integer.

Hence INSUFF.Hope this helps!
Then for X = 1/2

sqrt (3)*sqrt(1/2) = non integer

For x = 9

sqrt(3)*sqrt(9) = non integer

Both of these values would make Statement II sufficient.
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by cramya » Sat Dec 13, 2008 8:20 am
Then for X = 1/2
x is a positive integer[/list]
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